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Chapter 4: Problem 78

A \(46.2-\mathrm{mL}, 0.568 M\) calcium nitrate \(\left[\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\right]\) solution is mixed with \(80.5 \mathrm{~mL}\) of \(1.396 \mathrm{M}\) calcium nitrate solution. Calculate the concentration of the final solution.

Short Answer

Expert verified
The concentration of the final solution is 1.094 M

Step by step solution

01

Calculate the amount of moles in each solution

The formula for molarity \(M\) is \(\frac{moles}{L}\). To find the number of moles, you can multiply the molarity by the volume in liters. For the first solution, \(moles_1 = M1*V1 = 0.568 M * 0.0462 L = 0.02625 moles\). For the second solution, \(moles_2 = M2*V2 = 1.396 M * 0.0805 L = 0.11234 moles\).
02

Calculate the total volume of the final solution

The total volume of the final solution \(V_f\) is the sum of the volumes of the two solutions i.e. \(V_f = V1 + V2 = 0.0462 L + 0.0805 L = 0.1267 L\).
03

Calculate the total moles of solute in the final solution

The total moles of solute in the final solution obtained by combining the initial solutions is \(moles_{total} = moles_1 + moles_2 = 0.02625 moles+0.11234 moles = 0.13859 moles\).
04

Calculate the molarity of the final solution

The molarity of the final solution is calculated by dividing the total moles of solute by the total volume of the solution. So, \(M_f = \frac{moles_{total}}{V_f} = \frac{0.13859 moles}{0.1267 L} = 1.094 M\)

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Most popular questions from this chapter

Acetylsalicylic acid ( \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) ) is a monoprotic acid commonly known as "aspirin." A typical aspirin tablet, however, contains only a small amount of the acid. In an experiment to determine its composition, an aspirin tablet was crushed and dissolved in water. It took \(12.25 \mathrm{~mL}\) of \(0.1466 \mathrm{M} \mathrm{NaOH}\) to neutralize the solution. Calculate the number of grains of aspirin in the tablet. (One grain \(=0.0648\) g.)

Sulfites (compounds containing the \(\mathrm{SO}_{3}^{2-}\) ions) are used as preservatives in dried fruits and vegetables and in wine making. In an experiment to test the presence of sulfite in fruit, a student first soaked several dried apricots in water overnight and then filtered the solution to remove all solid particles. She then treated the solution with hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) to oxidize the sulfite ions to sulfate ions. Finally, the sulfate ions were precipitated by treating the solution with a few drops of a barium chloride \(\left(\mathrm{BaCl}_{2}\right)\) solution. Write a balanced equation for each of the preceding steps.

On the basis of oxidation number considerations, one of the following oxides would not react with molecular oxygen: \(\mathrm{NO}, \mathrm{N}_{2} \mathrm{O}, \mathrm{SO}_{2}, \mathrm{SO}_{3}, \mathrm{P}_{4} \mathrm{O}_{6}\) Which one is it? Why?

The following "cycle of copper" experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is reacted with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with \(65.6 \mathrm{~g}\) of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.

(a) Describe a preparation for magnesium hydroxide \(\left[\mathrm{Mg}(\mathrm{OH})_{2}\right]\) and predict its solubility. (b) Milk of magnesia contains mostly \(\mathrm{Mg}(\mathrm{OH})_{2}\) and is effective in treating acid (mostly hydrochloric acid) indigestion. Calculate the volume of a \(0.035 \mathrm{M} \mathrm{HCl}\) solution (a typical acid concentration in an upset stomach) needed to react with two spoonfuls (approximately \(10 \mathrm{~mL}\) ) of milk of magnesia [at \(0.080 \mathrm{~g}\) \(\left.\mathrm{Mg}(\mathrm{OH})_{2} / \mathrm{mL}\right]\).
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